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The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined - HSC - SSCE Chemistry - Question 32 - 2022 - Paper 1
Question 32
The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined. Step 1: A solution of NaOH(aq) was standardised by titrating it... show full transcript
Worked Solution & Example Answer:The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined - HSC - SSCE Chemistry - Question 32 - 2022 - Paper 1
Step 1
Titration 1
Answer
Amount of KHP = 4.989 g Molar mass of KHP = 204.22 g mol⁻¹ Moles of KHP = ( \frac{4.989 \text{ g}}{204.22 \text{ g mol}^{-1}} = 0.024495 \text{ moles} )
Volume of KHP solution = 25.00 mL = 0.02500 L Concentration of KHP = ( \frac{0.024495 \text{ moles}}{0.02500 \text{ L}} = 0.9798 \text{ mol L}^{-1} )
Volume NaOH (average) = 27.40 mL = 0.02740 L Moles of NaOH = Concentration of KHP x Volume of KHP = 0.024495 x 0.02740 = 0.0006722 moles
Concentration of NaOH = ( \frac{0.0006722}{0.02740} = 0.0245 \text{ mol L}^{-1} )
Step 2
Titration 2
Answer
Volume NaOH = 13.10 mL = 0.01310 L Concentration of NaOH = 0.2228972333 mol L⁻¹
Amount of NaOH = 0.2228972333 mol L⁻¹ x 0.01310 L = 0.00291995375 moles
Volume of citric acid = 25.00 mL = 0.02500 L The ratio of citric acid to NaOH is 1:3 Amount of citric acid = Moles NaOH ÷ 3 = 0.00097331791 moles
Concentration of citric acid = ( \frac{0.00097331791}{0.02500} = 0.0389 \text{ mol L}^{-1} )